How to find the extension of an ideal
I have a very elementary question and I would appreciate understanding.
Let $A$ be a commutative and unitary ring, and $I$ an ideal of this ring.
Considering the inclusion $i: Alongrightarrow A[X]$, I need to prove that $I^e=I[X]$, where $I^e$ is the extension of the ideal $I$.
I know this is true because I have sometimes used it in other problems, but I actually don't know the reasoning behind it.
Thanks! Any help or explanation would be appreciated.
abstract-algebra ring-theory ideals
add a comment |
I have a very elementary question and I would appreciate understanding.
Let $A$ be a commutative and unitary ring, and $I$ an ideal of this ring.
Considering the inclusion $i: Alongrightarrow A[X]$, I need to prove that $I^e=I[X]$, where $I^e$ is the extension of the ideal $I$.
I know this is true because I have sometimes used it in other problems, but I actually don't know the reasoning behind it.
Thanks! Any help or explanation would be appreciated.
abstract-algebra ring-theory ideals
What do you mean by $I^e$?
– Alex Vong
Nov 25 at 20:24
2
The extension of the ideal I
– S. Prevč
Nov 25 at 20:25
I forget to read the title...
– Alex Vong
Nov 25 at 20:25
By definition, $I^e$ is the ideal generated by $i(I) = I$, so $I^e = {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$. Now we only need to show ${a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]} = I[x]$, can you show it?
– Alex Vong
Nov 25 at 20:58
I can see one inclusion ($<i(I)>$ included in $I[X]$), but not the other one.
– S. Prevč
Nov 25 at 21:13
add a comment |
I have a very elementary question and I would appreciate understanding.
Let $A$ be a commutative and unitary ring, and $I$ an ideal of this ring.
Considering the inclusion $i: Alongrightarrow A[X]$, I need to prove that $I^e=I[X]$, where $I^e$ is the extension of the ideal $I$.
I know this is true because I have sometimes used it in other problems, but I actually don't know the reasoning behind it.
Thanks! Any help or explanation would be appreciated.
abstract-algebra ring-theory ideals
I have a very elementary question and I would appreciate understanding.
Let $A$ be a commutative and unitary ring, and $I$ an ideal of this ring.
Considering the inclusion $i: Alongrightarrow A[X]$, I need to prove that $I^e=I[X]$, where $I^e$ is the extension of the ideal $I$.
I know this is true because I have sometimes used it in other problems, but I actually don't know the reasoning behind it.
Thanks! Any help or explanation would be appreciated.
abstract-algebra ring-theory ideals
abstract-algebra ring-theory ideals
edited Nov 25 at 20:25
asked Nov 25 at 20:15
S. Prevč
12
12
What do you mean by $I^e$?
– Alex Vong
Nov 25 at 20:24
2
The extension of the ideal I
– S. Prevč
Nov 25 at 20:25
I forget to read the title...
– Alex Vong
Nov 25 at 20:25
By definition, $I^e$ is the ideal generated by $i(I) = I$, so $I^e = {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$. Now we only need to show ${a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]} = I[x]$, can you show it?
– Alex Vong
Nov 25 at 20:58
I can see one inclusion ($<i(I)>$ included in $I[X]$), but not the other one.
– S. Prevč
Nov 25 at 21:13
add a comment |
What do you mean by $I^e$?
– Alex Vong
Nov 25 at 20:24
2
The extension of the ideal I
– S. Prevč
Nov 25 at 20:25
I forget to read the title...
– Alex Vong
Nov 25 at 20:25
By definition, $I^e$ is the ideal generated by $i(I) = I$, so $I^e = {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$. Now we only need to show ${a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]} = I[x]$, can you show it?
– Alex Vong
Nov 25 at 20:58
I can see one inclusion ($<i(I)>$ included in $I[X]$), but not the other one.
– S. Prevč
Nov 25 at 21:13
What do you mean by $I^e$?
– Alex Vong
Nov 25 at 20:24
What do you mean by $I^e$?
– Alex Vong
Nov 25 at 20:24
2
2
The extension of the ideal I
– S. Prevč
Nov 25 at 20:25
The extension of the ideal I
– S. Prevč
Nov 25 at 20:25
I forget to read the title...
– Alex Vong
Nov 25 at 20:25
I forget to read the title...
– Alex Vong
Nov 25 at 20:25
By definition, $I^e$ is the ideal generated by $i(I) = I$, so $I^e = {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$. Now we only need to show ${a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]} = I[x]$, can you show it?
– Alex Vong
Nov 25 at 20:58
By definition, $I^e$ is the ideal generated by $i(I) = I$, so $I^e = {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$. Now we only need to show ${a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]} = I[x]$, can you show it?
– Alex Vong
Nov 25 at 20:58
I can see one inclusion ($<i(I)>$ included in $I[X]$), but not the other one.
– S. Prevč
Nov 25 at 21:13
I can see one inclusion ($<i(I)>$ included in $I[X]$), but not the other one.
– S. Prevč
Nov 25 at 21:13
add a comment |
2 Answers
2
active
oldest
votes
To see the other inclusion, take any $f in I[x]$, then $f = a_0 + a_1 x + dots a_n x^n$ for some $a_j in I$. Now notice that $a_j in I$ and $x^j in A[x]$, so we have $f in {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$.
add a comment |
Alex Vong gave a good suggestion in the comments, but there's another way to see this. Hopefully this answer is helpful, but to be honest, your answer is missing context (by which I mean, you haven't told us what you've tried or what your thoughts are), so when I answer such questions, it's mostly for fun, since without context, I can't know what will be most useful to you.
Much of this answer is exposition. If you're familiar with all the material contained here, this argument is much shorter.
The extension of an ideal satisfies a universal property. Let $f:Ato B$ be a ring homomorphism, $I$ an ideal of $A$, then $I^e:=f(I)B$ is the smallest ideal of $B$ containing the image of $I$, which means that if $g:Bto C$ is any ring homomorphism such that $f(I)subseteq ker g$, we must have $I^esubseteq ker g$. $I^e$ is the unique ideal with this property, since we can take $g$ to be the quotient $Bto B/I^e$, and then if another ideal $J$ satisfied the property, we would have $f(I)subseteq Jsubseteq ker g= I^e$, and $I^e$ is the smallest ideal of $B$ containing $I^e$, so $J=I^e$.
Now the first isomorphism theorem says that if $g:Bto C$ then $I^esubseteq ker g$ is equivalent to the existence of a map $tilde{g} : B/I^e to C$ such that $tilde{g}circ q = g$, where $q:Bto B/I^e$ is the quotient.
Combining these pieces of information, we get the following.
If $A$ is a commutative, unital ring, and $Isubseteq A$ is an ideal. Then the extension of $I$ along the structure map $iota:Ato A[x]$ is the unique ideal $I^e$ such that for all maps $g$ from $A[x]to C$ such that $Isubseteq ker g$, there exists a map $tilde{g}:A[x]/I^e to C$ with $tilde{g}circ q = g$.
Now we recall the universal properties of polynomial rings. Maps $g:A[x]to C$ correspond precisely to pairs $(g',c)$ with $g'$ a map $g' : Ato C$ and $c$ an element of $C$. Maps $g:A[x]to C$ with $Isubseteq ker g$ therefore correspond to pairs $(g',c)$ with $Isubseteq ker g'$. But by the universal property of the quotient, these correspond precisely to pairs $(bar{g},c)$ with $bar{g}$ a map from $A/Ito C$ and $cin C$. Thus maps from $A[x]to C$ which annihilate $I$ correspond precisely to maps from $A/I[x]$ to $C$. Conveniently, we have a surjective map $A[x]to A/I[x]$ whose kernel is precisely $I[x]$ (this part is obvious from the definition of the map from $A[x]to A/I[x]$, which is reduce every coefficient mod $I$). Thus by the universal property (of sorts) of $I^e$, since $A[x]/I[x]$ has the property characterizing $I^e$, $I[x]=I^e$.
add a comment |
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To see the other inclusion, take any $f in I[x]$, then $f = a_0 + a_1 x + dots a_n x^n$ for some $a_j in I$. Now notice that $a_j in I$ and $x^j in A[x]$, so we have $f in {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$.
add a comment |
To see the other inclusion, take any $f in I[x]$, then $f = a_0 + a_1 x + dots a_n x^n$ for some $a_j in I$. Now notice that $a_j in I$ and $x^j in A[x]$, so we have $f in {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$.
add a comment |
To see the other inclusion, take any $f in I[x]$, then $f = a_0 + a_1 x + dots a_n x^n$ for some $a_j in I$. Now notice that $a_j in I$ and $x^j in A[x]$, so we have $f in {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$.
To see the other inclusion, take any $f in I[x]$, then $f = a_0 + a_1 x + dots a_n x^n$ for some $a_j in I$. Now notice that $a_j in I$ and $x^j in A[x]$, so we have $f in {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$.
answered Nov 25 at 21:40
Alex Vong
1,292819
1,292819
add a comment |
add a comment |
Alex Vong gave a good suggestion in the comments, but there's another way to see this. Hopefully this answer is helpful, but to be honest, your answer is missing context (by which I mean, you haven't told us what you've tried or what your thoughts are), so when I answer such questions, it's mostly for fun, since without context, I can't know what will be most useful to you.
Much of this answer is exposition. If you're familiar with all the material contained here, this argument is much shorter.
The extension of an ideal satisfies a universal property. Let $f:Ato B$ be a ring homomorphism, $I$ an ideal of $A$, then $I^e:=f(I)B$ is the smallest ideal of $B$ containing the image of $I$, which means that if $g:Bto C$ is any ring homomorphism such that $f(I)subseteq ker g$, we must have $I^esubseteq ker g$. $I^e$ is the unique ideal with this property, since we can take $g$ to be the quotient $Bto B/I^e$, and then if another ideal $J$ satisfied the property, we would have $f(I)subseteq Jsubseteq ker g= I^e$, and $I^e$ is the smallest ideal of $B$ containing $I^e$, so $J=I^e$.
Now the first isomorphism theorem says that if $g:Bto C$ then $I^esubseteq ker g$ is equivalent to the existence of a map $tilde{g} : B/I^e to C$ such that $tilde{g}circ q = g$, where $q:Bto B/I^e$ is the quotient.
Combining these pieces of information, we get the following.
If $A$ is a commutative, unital ring, and $Isubseteq A$ is an ideal. Then the extension of $I$ along the structure map $iota:Ato A[x]$ is the unique ideal $I^e$ such that for all maps $g$ from $A[x]to C$ such that $Isubseteq ker g$, there exists a map $tilde{g}:A[x]/I^e to C$ with $tilde{g}circ q = g$.
Now we recall the universal properties of polynomial rings. Maps $g:A[x]to C$ correspond precisely to pairs $(g',c)$ with $g'$ a map $g' : Ato C$ and $c$ an element of $C$. Maps $g:A[x]to C$ with $Isubseteq ker g$ therefore correspond to pairs $(g',c)$ with $Isubseteq ker g'$. But by the universal property of the quotient, these correspond precisely to pairs $(bar{g},c)$ with $bar{g}$ a map from $A/Ito C$ and $cin C$. Thus maps from $A[x]to C$ which annihilate $I$ correspond precisely to maps from $A/I[x]$ to $C$. Conveniently, we have a surjective map $A[x]to A/I[x]$ whose kernel is precisely $I[x]$ (this part is obvious from the definition of the map from $A[x]to A/I[x]$, which is reduce every coefficient mod $I$). Thus by the universal property (of sorts) of $I^e$, since $A[x]/I[x]$ has the property characterizing $I^e$, $I[x]=I^e$.
add a comment |
Alex Vong gave a good suggestion in the comments, but there's another way to see this. Hopefully this answer is helpful, but to be honest, your answer is missing context (by which I mean, you haven't told us what you've tried or what your thoughts are), so when I answer such questions, it's mostly for fun, since without context, I can't know what will be most useful to you.
Much of this answer is exposition. If you're familiar with all the material contained here, this argument is much shorter.
The extension of an ideal satisfies a universal property. Let $f:Ato B$ be a ring homomorphism, $I$ an ideal of $A$, then $I^e:=f(I)B$ is the smallest ideal of $B$ containing the image of $I$, which means that if $g:Bto C$ is any ring homomorphism such that $f(I)subseteq ker g$, we must have $I^esubseteq ker g$. $I^e$ is the unique ideal with this property, since we can take $g$ to be the quotient $Bto B/I^e$, and then if another ideal $J$ satisfied the property, we would have $f(I)subseteq Jsubseteq ker g= I^e$, and $I^e$ is the smallest ideal of $B$ containing $I^e$, so $J=I^e$.
Now the first isomorphism theorem says that if $g:Bto C$ then $I^esubseteq ker g$ is equivalent to the existence of a map $tilde{g} : B/I^e to C$ such that $tilde{g}circ q = g$, where $q:Bto B/I^e$ is the quotient.
Combining these pieces of information, we get the following.
If $A$ is a commutative, unital ring, and $Isubseteq A$ is an ideal. Then the extension of $I$ along the structure map $iota:Ato A[x]$ is the unique ideal $I^e$ such that for all maps $g$ from $A[x]to C$ such that $Isubseteq ker g$, there exists a map $tilde{g}:A[x]/I^e to C$ with $tilde{g}circ q = g$.
Now we recall the universal properties of polynomial rings. Maps $g:A[x]to C$ correspond precisely to pairs $(g',c)$ with $g'$ a map $g' : Ato C$ and $c$ an element of $C$. Maps $g:A[x]to C$ with $Isubseteq ker g$ therefore correspond to pairs $(g',c)$ with $Isubseteq ker g'$. But by the universal property of the quotient, these correspond precisely to pairs $(bar{g},c)$ with $bar{g}$ a map from $A/Ito C$ and $cin C$. Thus maps from $A[x]to C$ which annihilate $I$ correspond precisely to maps from $A/I[x]$ to $C$. Conveniently, we have a surjective map $A[x]to A/I[x]$ whose kernel is precisely $I[x]$ (this part is obvious from the definition of the map from $A[x]to A/I[x]$, which is reduce every coefficient mod $I$). Thus by the universal property (of sorts) of $I^e$, since $A[x]/I[x]$ has the property characterizing $I^e$, $I[x]=I^e$.
add a comment |
Alex Vong gave a good suggestion in the comments, but there's another way to see this. Hopefully this answer is helpful, but to be honest, your answer is missing context (by which I mean, you haven't told us what you've tried or what your thoughts are), so when I answer such questions, it's mostly for fun, since without context, I can't know what will be most useful to you.
Much of this answer is exposition. If you're familiar with all the material contained here, this argument is much shorter.
The extension of an ideal satisfies a universal property. Let $f:Ato B$ be a ring homomorphism, $I$ an ideal of $A$, then $I^e:=f(I)B$ is the smallest ideal of $B$ containing the image of $I$, which means that if $g:Bto C$ is any ring homomorphism such that $f(I)subseteq ker g$, we must have $I^esubseteq ker g$. $I^e$ is the unique ideal with this property, since we can take $g$ to be the quotient $Bto B/I^e$, and then if another ideal $J$ satisfied the property, we would have $f(I)subseteq Jsubseteq ker g= I^e$, and $I^e$ is the smallest ideal of $B$ containing $I^e$, so $J=I^e$.
Now the first isomorphism theorem says that if $g:Bto C$ then $I^esubseteq ker g$ is equivalent to the existence of a map $tilde{g} : B/I^e to C$ such that $tilde{g}circ q = g$, where $q:Bto B/I^e$ is the quotient.
Combining these pieces of information, we get the following.
If $A$ is a commutative, unital ring, and $Isubseteq A$ is an ideal. Then the extension of $I$ along the structure map $iota:Ato A[x]$ is the unique ideal $I^e$ such that for all maps $g$ from $A[x]to C$ such that $Isubseteq ker g$, there exists a map $tilde{g}:A[x]/I^e to C$ with $tilde{g}circ q = g$.
Now we recall the universal properties of polynomial rings. Maps $g:A[x]to C$ correspond precisely to pairs $(g',c)$ with $g'$ a map $g' : Ato C$ and $c$ an element of $C$. Maps $g:A[x]to C$ with $Isubseteq ker g$ therefore correspond to pairs $(g',c)$ with $Isubseteq ker g'$. But by the universal property of the quotient, these correspond precisely to pairs $(bar{g},c)$ with $bar{g}$ a map from $A/Ito C$ and $cin C$. Thus maps from $A[x]to C$ which annihilate $I$ correspond precisely to maps from $A/I[x]$ to $C$. Conveniently, we have a surjective map $A[x]to A/I[x]$ whose kernel is precisely $I[x]$ (this part is obvious from the definition of the map from $A[x]to A/I[x]$, which is reduce every coefficient mod $I$). Thus by the universal property (of sorts) of $I^e$, since $A[x]/I[x]$ has the property characterizing $I^e$, $I[x]=I^e$.
Alex Vong gave a good suggestion in the comments, but there's another way to see this. Hopefully this answer is helpful, but to be honest, your answer is missing context (by which I mean, you haven't told us what you've tried or what your thoughts are), so when I answer such questions, it's mostly for fun, since without context, I can't know what will be most useful to you.
Much of this answer is exposition. If you're familiar with all the material contained here, this argument is much shorter.
The extension of an ideal satisfies a universal property. Let $f:Ato B$ be a ring homomorphism, $I$ an ideal of $A$, then $I^e:=f(I)B$ is the smallest ideal of $B$ containing the image of $I$, which means that if $g:Bto C$ is any ring homomorphism such that $f(I)subseteq ker g$, we must have $I^esubseteq ker g$. $I^e$ is the unique ideal with this property, since we can take $g$ to be the quotient $Bto B/I^e$, and then if another ideal $J$ satisfied the property, we would have $f(I)subseteq Jsubseteq ker g= I^e$, and $I^e$ is the smallest ideal of $B$ containing $I^e$, so $J=I^e$.
Now the first isomorphism theorem says that if $g:Bto C$ then $I^esubseteq ker g$ is equivalent to the existence of a map $tilde{g} : B/I^e to C$ such that $tilde{g}circ q = g$, where $q:Bto B/I^e$ is the quotient.
Combining these pieces of information, we get the following.
If $A$ is a commutative, unital ring, and $Isubseteq A$ is an ideal. Then the extension of $I$ along the structure map $iota:Ato A[x]$ is the unique ideal $I^e$ such that for all maps $g$ from $A[x]to C$ such that $Isubseteq ker g$, there exists a map $tilde{g}:A[x]/I^e to C$ with $tilde{g}circ q = g$.
Now we recall the universal properties of polynomial rings. Maps $g:A[x]to C$ correspond precisely to pairs $(g',c)$ with $g'$ a map $g' : Ato C$ and $c$ an element of $C$. Maps $g:A[x]to C$ with $Isubseteq ker g$ therefore correspond to pairs $(g',c)$ with $Isubseteq ker g'$. But by the universal property of the quotient, these correspond precisely to pairs $(bar{g},c)$ with $bar{g}$ a map from $A/Ito C$ and $cin C$. Thus maps from $A[x]to C$ which annihilate $I$ correspond precisely to maps from $A/I[x]$ to $C$. Conveniently, we have a surjective map $A[x]to A/I[x]$ whose kernel is precisely $I[x]$ (this part is obvious from the definition of the map from $A[x]to A/I[x]$, which is reduce every coefficient mod $I$). Thus by the universal property (of sorts) of $I^e$, since $A[x]/I[x]$ has the property characterizing $I^e$, $I[x]=I^e$.
answered Nov 25 at 22:32
jgon
12.4k21940
12.4k21940
add a comment |
add a comment |
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What do you mean by $I^e$?
– Alex Vong
Nov 25 at 20:24
2
The extension of the ideal I
– S. Prevč
Nov 25 at 20:25
I forget to read the title...
– Alex Vong
Nov 25 at 20:25
By definition, $I^e$ is the ideal generated by $i(I) = I$, so $I^e = {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$. Now we only need to show ${a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]} = I[x]$, can you show it?
– Alex Vong
Nov 25 at 20:58
I can see one inclusion ($<i(I)>$ included in $I[X]$), but not the other one.
– S. Prevč
Nov 25 at 21:13